1. Radius of circular motion in a magnetic field

For a charged particle moving perpendicular to a magnetic field, the magnetic force provides the centripetal force:

$qvB = \frac{mv^2}{r}$

So,

$r = \frac{mv}{qB} = \frac{p}{qB}$

where $p = mv$ is the momentum.

2. Use the fact that all have the same kinetic energy

Given same kinetic energy $K$,

$K = \frac{p^2}{2m}$

Hence,

$p = \sqrt{2mK}$

Substitute into radius formula:

$r = \frac{\sqrt{2mK}}{qB}$

Since $K$ and $B$ are same for all particles,

$r \propto \frac{\sqrt{m}}{q}$

So we compare $\dfrac{\sqrt{m}}{q}$ for each particle.

3. For each particle

• Electron: mass $m_e$, charge magnitude $e$

$r_e \propto \frac{\sqrt{m_e}}{e}$

• Proton: mass $m_p$, charge $e$

$r_p \propto \frac{\sqrt{m_p}}{e}$

Since $m_p \gg m_e$,

$r_p > r_e$

• Helium nucleus ($\alpha$-particle): mass approximately $4m_p$, charge $2e$

$r_{\mathrm{He}} \propto \frac{\sqrt{4m_p}}{2e} = \frac{2\sqrt{m_p}}{2e} = \frac{\sqrt{m_p}}{e}$

Thus,

$r_{\mathrm{He}} = r_p$

4. Final comparison

Therefore,

$r_e < r_p = r_{\mathrm{He}}$

So the correct option is B.

5. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They agree.